GALOIS GROUPS AS PERMUTATION GROUPS KEITH CONRAD 1. Introduction A Galois group is a group of field automorphisms under composition. By looking at the effect of a Galois group on field generators we can interpret the Galois group as permu- tations, which makes it a subgroup of a symmetric group. This makes Galois groups into relatively concrete objects and is particularly effective when the Galois group turns out to be a symmetric or alternating group. 2. Automorphisms of fields as permutations of roots The Galois group of a polynomial f(T ) 2 K[T ] over K is defined to be the Galois group of a splitting field for f(T ) over K. We do not require f(T ) to be irreducible in K[T ]. p Example 2.1. The polynomial T 4 − 2 has splitting field Q( 4 2; i) over Q, so the Galois of 4 4 T − 2 over Q is isomorphic to D4. The splitting field of T − 2 over R is C, so the Galois group of T 4 − 2 over R is Gal(C=R) = fz 7! z; z 7! zg, which is cyclic of order 2. Example 2.2. The Galois group of (T 2 − 2)(T 2 − 3) over Q is isomorphic to Z=2Z × Z=2Z. Its Galois group over R is trivial since the polynomial splits completely over R. Writing f(T ) = (T − r1) ··· (T − rn), the splitting field of f(T ) over K is K(r1; : : : ; rn). Each σ in the Galois group of f(T ) over K permutes the ri's since σ fixes K and therefore f(r) = 0 ) f(σ(r)) = 0. The automorphism σ is completely determined by its permutation of the ri's since the ri's generate the splitting field over K. A permutation of the ri's can be viewed as a permutation of the subscripts 1; 2; : : : ; n. 4 Examplep p 2.3.p Considerp the Galois group of T p− 2 over Q. The polynomial has 4 roots: 4 4 4 4 4 2; i 2;p− 2; −pi 2. Two generators ofp Gal(Qp( 2; i)=Q) are the automorphisms r and s 4 4 4 4 where r( p2) = i 2 and r(i) = i, and s( 2) = 2 and s(i) = −i. The effect of the Galois group on 4 2 and i is in Table1. Automorphism 1 r r2 r3 s rs r2s r3s p p p p p p p p p Value on 4 2 4 2 i 4 2 − 4 2 −i 4 2 4 2 i 4 2 − 4 2 −i 4 2 Value on i i i i i −i −i −i −i Table 1. The effect of r on the roots of T 4 − 2 is p p p p p p p p r( 4 2) = i 4 2; r(i 4 2) = − 4 2; r(− 4 2) = −i 4 2; r(−i 4 2) = 4 2; which is a 4-cycle, while the effect of s on the roots of T 4 − 2 is p p p p p p p p s( 4 2) = 4 2; s(i 4 2) = −i 4 2; s(−i 4 2) = i 4 2; s(− 4 2) = − 4 2; 1 2 KEITH CONRAD p p p p which swaps i 4 2 and −i 4 2 while fixing 4 2 and − 4 2. So s is a 2-cycle on the roots. Indexing the roots of T 4 − 2 as p4 p4 p4 p4 (2.1) r1 = 2; r2 = i 2; r3 = − 2; r4 = −i 2; the automorphism r acts on the roots like (1234) and the automorphism s acts on the roots like (24). With this indexing of the roots, the Galois group of T 4 − 2 over Q becomes the group of permutations in S4 in Table2. Automorphism 1 r r2 r3 s rs r2s r3s Permutation (1) (1234) (13)(24) (1432) (24) (12)(34) (13) (14)(23) Table 2. Example 2.4. If we label the roots of (T 2 − 2)(T 2 − 3) as p p p p r1 = 2; r2 = − 2; r3 = 3; r4 = − 3; 2 2 then the Galois group of (T − 2)(T − 3) over Q becomes the following subgroup of S4: (2.2) (1); (12); (34); (12)(34): Numbering the roots of f(T ) in different ways can identify the Galois group with different subgroups of Sn. p p p p 4 4 4 4 Example 2.5. Renaming 2; i 2; − 2; −i 2 in this order as r2; r4; r3; r1 identifies the 4 Galois group of T − 2 over Q with the subgroup of S4 in Table3, which is not the same subgroup of S4 in Table2. Automorphism 1 r r2 r3 s rs r2s r3s Permutation (1) (1243) (14)(23) (1342) (14) (13)(24) (23) (12)(34) Table 3. p p p p Example 2.6. If we label 2; − 2; 3; − 3 in this order as r2; r4; r1; r3 then the Galois 2 2 group of (T − 2)(T − 3) over Q turns into the following subgroup of S4: (2.3) (1); (13); (24); (13)(24): This is not the same subgroup as (2.2). In general, associating to each σ in the Galois group of f(T ) over K its permutation on the roots of f(T ), viewed as a permutation of the subscripts of the roots when we list them as r1; : : : ; rn, is a homomorphism from the Galois group to Sn. This homomorphism is injective since its kernel is trivial: an element of the Galois group that fixes each ri is the identity on the splitting field. Thinking about the Galois group of a polynomial with degree n as a subgroup of Sn is the original viewpoint of Galois. (The description of Galois theory in terms of field automorphisms is due to Dedekind and, with more abstraction, Artin.) Two different choices for indexing the roots of f(T ) can lead to different subgroups of Sn, but they will be conjugate subgroups. For instance, the subgroups in Tables2 and 1234 3 are conjugate by the permutation 2431 = (124), which is the permutation turning one indexing of the roots into the other, and the subgroups (2.2) and (2.3) are conjugate by GALOIS GROUPS AS PERMUTATION GROUPS 3 1234 2413 = (1243). Although the Galois group of f(T ) over K does not have a canonical embedding into Sn in general, its image in Sn is well-defined up to an overall conjugation. For example, without fixing an indexing of the roots, it doesn't make sense to ask if a particular permutation like (132) is in the Galois group as a subgroup of Sn, but it does make sense to ask if the Galois group contains a permutation with a particular cycle type (like a 3-cycle). We can speak about Galois groups of irreducible or reducible polynomials, like T 4 − 2 or (T 2 − 2)(T 3 − 2) over Q. Only for an irreducible polynomial does the Galois group have a special property, called transitivity, when we turn the Galois group into a subgroup of Sn. A subgroup G ⊂ Sn is called transitive when, for all i 6= j in f1; 2; : : : ; ng, there is a permutation in G sending i to j. Example 2.7. The subgroups of S4 in Tables2 and3 are transitive. This corresponds to the fact that for each pair of roots of T 4 − 2 there is an element of its Galois group over Q taking the first root to the second. Example 2.8. The subgroup of S4 in (2.2) is not transitive since no elementp p of the subgroup ptakes 1p to 3. This corresponds to the fact that an element of Gal(Q( 2; 3)=Q) can't send 2 to 3. Being transitive is not a property of an abstract group. It is a property of subgroups of 1 Sn. A conjugate subgroup of a transitive subgroup of Sn is also transitive since conjugation on Sn amounts to listing the numbers from 1 to n in a different order. Theorem 2.9. Let f(T ) 2 K[T ] be a separable polynomial of degree n. (a) If f(T ) is irreducible in K[T ] then its Galois group over K has order divisible by n. (b) The polynomial f(T ) is irreducible in K[T ] if and only if its Galois group over K is a transitive subgroup of Sn. Proof. (a) For a root r of f(T ) in K,[K(r): K] = n is a factor of the degree of the splitting field over K, which is the size of the Galois group over K. (b) First suppose f(T ) is irreducible. For two roots ri and rj of f(T ), we can write rj = σ(ri) for some σ in the Galois group of f(T ) over K. Therefore the Galois group, as a subgroup of Sn, sends i to j, so it is a transitive subgroup. Now suppose f(T ) is reducible (so n ≥ 2). It is a product of distinct irreducibles since it is separable. Let ri and rj be roots of different irreducible factors of f(T ). These irreducible factors are the minimal polynomials of ri and rj over K. For each σ in the Galois group of f(T ) over K, σ(ri) has the same minimal polynomial over K as ri, so we can't have σ(ri) = rj. Therefore, as a subgroup of Sn, the Galois group of f(T ) does not send i to j, so it is not a transitive subgroup of Sn.